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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

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96 Chapter 2 Limits and Derivatives

Therefore we have

lim f f sxd 1 5tsxdg − lim f sxd 1 lim f5tsxdg (by Limit Law 1)

x l 22 x l 22 x l 22

− lim f sxd 1 5 lim tsxd (by Limit Law 3)

x l 22 x l 22

− 1 1 5s21d − 24

(b) We see that lim x l 1 f sxd − 2. But lim x l 1 tsxd does not exist because the left and

right limits are different:

lim tsxd − 22 lim

x l 12 tsxd − 21

x l 11 So we can’t use Law 4 for the desired limit. But we can use Law 4 for the one-sided

limits:

lim

x l 1

lim

x l 1

f f sxdtsxdg − lim f sxd ? lim tsxd − 2 ? s22d − 24

2 2 2

x l1

x l1

f f sxdtsxdg − lim f sxd ? lim tsxd − 2 ? s21d − 22

1 1 1

x l1

The left and right limits aren’t equal, so lim x l 1 f f sxdtsxdg does not exist.

(c) The graphs show that

lim

x l 2

x l1

f sxd < 1.4 and lim

x l 2 tsxd − 0

Because the limit of the denominator is 0, we can’t use Law 5. The given limit does not

exist because the denominator approaches 0 while the numerator approaches a nonzero

number.

If we use the Product Law repeatedly with tsxd − f sxd, we obtain the following law.

Power Law

6. lim

x la f f sxdgn −

f lim

x la f sxd g n

where n is a positive integer

In applying these six limit laws, we need to use two special limits:

7. lim

x l a

c − c 8. lim

x l a

x − a

These limits are obvious from an intuitive point of view (state them in words or draw

graphs of y − c and y − x), but proofs based on the precise definition are requested in

the exercises for Section 2.4.

If we now put f sxd − x in Law 6 and use Law 8, we get another useful special limit.

9. lim

x l a

x n − a n

where n is a positive integer

A similar limit holds for roots as follows. (For square roots the proof is outlined in

Exercise 2.4.37.)

10. lim s n x − s n a where n is a positive integer

x l a

(If n is even, we assume that a . 0.)

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